(2/y-3)=-(7/4y-12)+2

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Solution for (2/y-3)=-(7/4y-12)+2 equation:


D( y )

y = 0

y = 0

y = 0

y in (-oo:0) U (0:+oo)

2/y-3 = 2-((7/4)*y)+12 // - 2-((7/4)*y)+12

(7/4)*y+2/y-12-3-2 = 0

7/4*y^1+2*y^-1-17*y^0 = 0

(7/4*y^2-17*y^1+2*y^0)/(y^1) = 0 // * y^2

y^1*(7/4*y^2-17*y^1+2*y^0) = 0

y^1

(7/4)*y^2-17*y+2 = 0

(7/4)*y^2-17*y+2 = 0

DELTA = (-17)^2-(2*4*(7/4))

DELTA = 275

DELTA > 0

y = (275^(1/2)+17)/(2*(7/4)) or y = (17-275^(1/2))/(2*(7/4))

y = 2/7*(5*11^(1/2)+17) or y = 2/7*(17-5*11^(1/2))

y in { 2/7*(17-5*11^(1/2)), 2/7*(5*11^(1/2)+17)}

y in { 2/7*(17-5*11^(1/2)), 2/7*(5*11^(1/2)+17) }

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